Ask question

Ask question

All categories

4 years ago

6

Which is a chemical reaction? Select one: a. a peanut shell cracking open into two halves b. solid chocolate melting into liquid

chocolate c. water vapor condensing into liquid water on a window d. hydrogen gas and oxygen gas synthesizing into liquid water

2 answers:

raketka [301]4 years ago

8 0

My best educated guess would be d

sveta [45]4 years ago

7 0

I agree ^^ I think it would be d

You might be interested in

What is the simulation theory?

Inessa05 [86]

The simulation hypothesis<span> contends that reality is in fact a simulation (most likely a </span>computer simulation<span>), of which we, the simulants, are totally unaware. Some versions rely on the development of simulated reality, a fictional technology.</span>

7 0

3 years ago

A 6 gram coin is which is initially at rest is dropped from the observation deck of a skyscraper 300 meters above the street bel

steposvetlana [31]

Answer:

a) The work done by gravity on the coin as it falls is 17.653 joules.

b) The kinetic energy of the coin at a speed of 60 meters per second is 10.8 joules.

c) The work lost due to air resistance is 6.853 joules.

d) The average force exerted on the coin due to air resistance as it fell is 0.023 newtons.

Explanation:

a) We must remind that situation with Earth-coin system must be represented by Principle of Energy Conservation and Work Energy Theorem. According to this latter, work done by gravity equals to the change in gravitational potential energy:

$\Delta U_{g} = m\cdot g \cdot \Delta z$ (Eq. 1)

Where:

$\Delta U_{g}$ - Change in gravitational potential energy, measured in joules.

$m$ - Mass, measured in kilograms.

$g$ - Gravitational accelerations, measured in meters per square second.

$\Delta z$ - Height of the skyscraper, measured in meters.

If we know that $m = 0.006\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta z = 300\,m$ , the work done gravity on the coin is:

$\Delta U_{g} = (0.006\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (300\,m)$

$\Delta U_{g} = 17.653\,J$

The work done by gravity on the coin as it falls is 17.653 joules.

b) By definition of translation kinetic energy, we get the following model:

$K = \frac{1}{2}\cdot m\cdot v^{2}$

Where:

$K$ - Kinetic energy of the coin right before hitting the street, measured in joules.

$v$ - Speed of the coin, measured in meters per second.

If we get that $m = 0.006\,kg$ and $v = 60\,\frac{m}{s}$ , the kinetic energy at this speed is:

$K = \frac{1}{2}\cdot (0.006\,kg)\cdot \left(60\,\frac{m}{s} \right)^{2}$

$K = 10.8\,J$

The kinetic energy of the coin at a speed of 60 meters per second is 10.8 joules.

c) The work lost due to air resistance is obtained derived from Principle of Energy Conservation and Work-Energy Theorem:

$W_{lost} = \Delta U_{g}-K$ (Eq. 2)

Where $W_{lost}$ is the work lost due to air resistance, measured in joules.

If we know that $\Delta U_{g} = 17.653\,J$ and $K = 10.8\,J$ , the work lost due to air resistance is:

$W_{lost} = 17.653\,J-10.8\,J$

$W_{lost} = 6.853\,J$

The work lost due to air resistance is 6.853 joules.

d) The average force exerted on the coin due to air resistance can be determined by applying definition of work, as air resistance force was antiparallel to the displacement of the coin. That is:

$W_{lost} = F\cdot \Delta z$

$F = \frac{W_{lost}}{\Delta z}$ (Eq. 3)

Where $F$ is the average force exerted on the coin due to air resistance, measured in newtons.

If we know that $W_{lost} = 6.853\,J$ and $\Delta z = 300\,m$ , then the average force exerted on the coin is:

$F = \frac{6.853\,J}{300\,m}$

$F = 0.023\,N$

The average force exerted on the coin due to air resistance as it fell is 0.023 newtons.

5 0

3 years ago

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 390 N while the ot

romanna [79]

Answer:

The crate's coefficient of kinetic friction on the floor is 0.23.

Explanation:

Given that,

Mass of the crate, m = 300 kg

One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.

The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :

$F_1+F_2=\mu N\\\\\mu=\dfrac{F_1+F_2}{mg}\\\\\mu=\dfrac{390+320}{300\times 10}\\\\\mu=0.23$

So, the crate's coefficient of kinetic friction on the floor is 0.23.

6 0

3 years ago

A 934 kg object at rest has what momentum ?

AlekseyPX

Answer:

0

Explanation:

An object that is at rest and not moving will always have 0 momentum!

I hope this helped!

6 0

2 years ago

A car traveling at 80 kilometers per hour is passed by a second car going in the same direction at a constant speed. After 30 se

gulaghasi [49]

Answer:

= 38.89 m/s

Explanation:

speed of first car (u) = 80 km/h = 22.22 m/s

time (t) = 30 s

distance (d) = 500 m

find the speed of second car (v)

let the speed of the second car be represented as 'v'

We can get the speed of the second car from the formula net speed = distance between cars / time interval

where

distance between cars = 500 m
time interval = 30 s
net speed = v-22.22 (since both cars move in the same direction)

substituting the above into the formula we have

v-22.22 = 500 / 30

v = 500/30 + 22.22

v = 38.89 m/s

5 0

3 years ago